The Optical Transport Hierarchy (OTH) technique is a new generation of transport hierarchy developed after the Synchronous Digital Hierarchy (SDH)/Synchronous Optical Network (SONET). For the transfer of the data in the OTH, Telecommunication Standardization Sector of the International Telecommunication Union (ITU-T) G.709 recommendation has defined signals to perform different functions, such as Optical channel Data Unit (ODU) as a connection signal, Optical channel Transport Unit (OTU) as a transport signal, Optical channel Payload Unit (OPU) and Optical channel Data Tributary Unit Group (ODTUG).
The frame format of ODUk is shown in FIG. 1, where k=1, 2, 3. ODUk has 4×3824=16320 bytes in total, where k=1, 2, 3. The area in Columns 1-14 of Row 1 is the area reserved for Frame Alignment (FA) and OTUk Overhead (OTUk OH), the area in Columns 1-14 of Rows 2-4 is the ODUk OH area, and the other area in Rows 15-3824 is the OPUk area which has 4×3810 bytes. Bytes 1-6 of the ODUk frame are FA bytes.
ODUk may be adapted to OTUk by mapping, and the adaptation of the frame format of ODUk to OTUk is shown in FIG. 2. Through filling the FA area with FA bytes and OTUk OH area with the OTUk overhead bytes, and through adding Columns 3825-4080 (4×256 bytes) which are filled with OTUk Forward Error Correction (FEC) Reed-Solomon code (RS), the OTUk frame is obtained. The OTUk frame has 4×4080=15296 bytes in total and the 1-6 bytes are Frame Alignment bytes.
The bit rate of ODUk (k=1, 2, 3) as a connection signal at three levels is calculated as follows:ODU1:239/238×2.48832 Gbps=2.498775126 Gbps;ODU2:239/237×9.95328 Gbps=10.037273924 Gbps;ODU3:239/236×39.81312 Gbps=40.319218983 Gbps.
In other words, the bit rate of ODUk (k=1, 2, 3) is calculated on the basis of the formula of 239/(239−k)×“bit rate of n order Synchronous Transfer Mode (STM-N)”.
The bit rate of OTUk (k=1, 2, 3) as a transport signal at three levels is calculated as follows:OTU1:255/238×2.48832 Gbps=2.66605714285714 Gbps;OTU2:255/237×9.95328 Gbps=10.7092253164557 Gbps;OTU3:255/236×39.81312 Gbps=43.018413559322 Gbps.
In other words, the bit rate of OTUk (k=1, 2, 3) is calculated on the basis of the formula of 255/(239−k)×“bit rate of STM-N”.
However, the existing signal transport technologies through a backplane and cross-point device often cannot support the transport of serial signals such as high rate signal ODU2/OTU2 or ODU3/OTU3. For example, currently, for the asynchronous cross-point device maturely applied in the industry, the bit rate of its port is up to 3.6 Gbps, which can only support the asynchronous switching function of the ODU1 serial signal.
The OTU2/ODU2 signal at a bit rate level of 10 Gbps is usually split into parallel signals of 4 bits, and the OTU3/ODU3 signal at a bit rate range of 40 Gbps is usually split into parallel signals of 16 bits. Framing processing of the split parallel signals also should be performed to ensure that the sink can combine the split parallel signals to obtain the source signal by frame alignment.
Currently, before transported, the OTN frame is split into 4 channels with 16 bytes in each block. FIG. 3 is a schematic diagram illustrating how the OTU2 frame is split in the OTN. As shown in FIG. 3, the area shaded with skew lines is the frame header area including OTUk FA. The OTU2 frame is usually split into 4 channels directly in the sequence of bytes, i.e. bytes 1-4080 for channel 1, bytes 4081-8160 for channel 2, bytes 8161-12240 for channel 3, bytes 12241-16320 for channel 4. As shown in FIG. 3a, the frame header area is in channel 1. Therefore, the sink can not implement the frame alignment and obtain the source signal through restoring and combining. For the purpose of solving this problem, as shown in FIG. 3b, the sequences of channels are adjusted at the beginning of the split frames 2, 3 and 4 so that the frame headers can be located in channels 2, 3 and 4 respectively. Therefore, each channel has a frame header area for frame alignment, and frame alignment and combination at the sink can be performed normally.
In the conventional method for splitting, it is required that each frame can be split into 2i channels and each channel includes integral blocks, wherein the “i” is a natural number. For a calculation of the size of one frame, supposing that the size of the frame is F, the number of channels is C, the number of blocks included in each channel is B, a block includes S bytes, and that the block is larger than the frame area, the size of the) frame F is equal to C×B×S, i.e. F=C×B×S.
Therefore, there is a requirement for the size of the frame according to the conventional method. With respect to the ODUk/OTUk frame format, the frame alignment area has 6 bytes, and the minimum size of a block should be 6 bytes.
For example, with respect to the OTU2 frame format above, the size of the OTU2 frame is 16320 bytes; when the OTU2 frame is split into 4 channels (for the 10 Gbps level, 4 channels can be processed conveniently), the size of the OTU2 frame 16320=6 byte/block×4 channels×255, i.e., there are 255 blocks in each channel after an OTU2 frame is split.
With respect to the ODU2 frame format, the OTU2 FA bytes are set in the ODU2 frame, where the FA area is the shaded area and includes 6 bytes. The size of the ODU2 frame is 15296 bytes, which satisfies the equation of 15296=16 byte/block×4 channel×239, i.e., there are 239 blocks in each channel after a frame is split.
With respect to the OTU3 frame format, the size of the OTU3 frame is 16320 bytes, and the OTU3 frame is split into 16 channels, and the size of the OTU3 frame satisfies the expression of 16320=16 byte/block×16 channel×63.75, i.e., there are 63.75 blocks in each channel after a frame is split. Because 63.75 is not an integer, the splitting can not be implemented.
With respect to the OTU3 frame format, the size of the OTU3 frame is 15296 bytes, the OTU3 frame is split into 16 channels, then the size of the OTU3 frame satisfies the expression of 16320=16 byte/block×16 channel×63.75, i.e., there are 59.75 blocks in each channel after a frame is split. Because 59.75 is not an integer, the splitting can not be implemented.
It can be seen that the above conventional method for signal splitting can not be applicable to some frame formats in the OTN system. Therefore, the signal processing capacity of the OTN system is not perfect enough, which is a shortcoming of the OTN system. As a result, many functions of the OTN system can not be implemented in many cases, such as the case of the uniformed transport and switch of various signals.